If `f (x)=((x-1) (x-2))/((x-3)(x-4)),` then number of local extremas for `g (x)`, where `g (x) =f (|x|):` (a) 3 (b) 4 (c) 5 (d) none of these

Options

A. 3
B. 4
C. 5
D. None of these

Answer

To solve the problem, we need to analyze the function \( g(x) = f(|x|) \) where \( f(x) = \frac{(x-1)(x-2)}{(x-3)(x-4)} \).

### Step-by-step Solution:

1. **Understanding the Function \( f(x) \)**:
The function \( f(x) \) is a rational function defined as:
\[ f(x) = \frac{(x-1)(x-2)}{(x-3)(x-4)} \]
This function is a quotient of two polynomials, and it can have local extrema where its derivative \( f'(x) \) is zero or undefined.

2. **Finding the Derivative \( f'(x) \)**:
To find the local extrema of \( f(x) \), we need to find its derivative using the quotient rule:
\[ f'(x) = \frac{(g(x)h'(x) - h(x)g'(x))}{(h(x))^2} \]
where \( g(x) = (x-1)(x-2) \) and \( h(x) = (x-3)(x-4) \).

3. **Identifying Critical Points**:
Set \( f'(x) = 0 \) to find critical points. The critical points will be where \( g(x) = 0 \) or where \( h(x) \) is undefined. The roots of \( g(x) \) are \( x = 1 \) and \( x = 2 \), which gives us two critical points.

4. **Behavior at Asymptotes**:
The function \( f(x) \) is undefined at \( x = 3 \) and \( x = 4 \) (vertical asymptotes). These points will also affect the behavior of the function and the number of local extrema.

5. **Considering \( g(x) = f(|x|) \)**:
Since \( g(x) = f(|x|) \), we need to consider both \( x \) and \( -x \):
- For \( x > 0 \), \( g(x) = f(x) \).
- For \( x < 0 \), \( g(x) = f(-x) \).
- Note that \( f(-x) = f(|x|) \) will yield the same critical points as \( f(x) \) for positive \( x \).

6. **Even Function Property**:
Since \( g(x) = f(|x|) \), it is an even function. This means that if there are \( n \) local extrema for \( x > 0 \), there will also be \( n \) local extrema for \( x < 0 \). Additionally, there will be an extremum at \( x = 0 \).

7. **Counting Local Extrema**:
- For \( x > 0 \): We found 2 local extrema from \( f(x) \).
- For \( x < 0 \): By symmetry, we also have 2 local extrema.
- At \( x = 0 \): There is 1 local extremum.

Thus, the total number of local extrema is:
\[ 2 \text{ (for } x > 0\text{)} + 2 \text{ (for } x < 0\text{)} + 1 \text{ (at } x = 0\text{)} = 5 \]

### Final Answer:
The number of local extrema for \( g(x) \) is \( \boxed{5} \).